K. A. Saaifan, Jacobs University, Bremen
9. The RLC Circuit
The RLC circuits have a wide range of applications, including oscillators and
frequency filters
This chapter considers the responses of RLC circuits
The result is a second-order differential equation for any voltage or current of
interest
We consider the following analysis
The Natural Response of a Parallel RLC Circuit
The Natural Response of a Series RLC Circuit
The Complete (Natural and Step) Response of RLC Circuits
1
2
K. A. Saaifan, Jacobs University, Bremen
9.1 The Source-Free Parallel Circuit
Obtaining the differential equation for a parallel RLC circuit
Apply KCL
v
i tiC t=0
R L
v 1 t
dv
 ∫t v d t iL t0 C
=0
R L
dt
iL(t)
0
Differentiate both sides with respect to time
d2 v 1 d v 1
C

 v=0
2
R
d
t
L
dt
Two initial conditions
The capacitor voltage cannot change abruptly
−

vC 0 =vC 0=vC 0 =V 0
1
The inductor current cannot change abruptly
iL 0−iL 0iL 0=I 0
−I 0 −V 0 / R
d vC t
=
∣
dt t=0
C
2
iC 0=−iL 0−iR 0
V
=−I 0 − 0
R
d vC t
iC 0=C
dt ∣t=0
iC(t)
3
K. A. Saaifan, Jacobs University, Bremen
Solution of the differential equation
We assume the exponential form of the natural response is
st
vt=Ae
Substitute into the ordinary differential equation, we got the characteristic
equation of s determined by the circuit parameters
1
1
d2 v 1 d v 1
2
C s  s =0
C

 v=0
2
R
L
R
d
t
L
dt
The characteristic equation has two roots
1
1
2
2
C
s

s
 v=0
1
1
1
R
L
s1,2 =−
±
−
LC
2 RC
2 RC


Thus, the natural response has the following form
s1 t
s2 t
vt=A1 e A 2 e
where the constants A1 and A2 are determined using the initial conditions
Definition of frequency terms
The resonant frequency
0 =
1
 LC
The damping coefficient
=
1
2RC
4
K. A. Saaifan, Jacobs University, Bremen
The roots of the characteristic equation can be expressed as
s1 =−   2−20
s2=−−  2 −20
Three types of natural response
Response
Criteria
Solutions
Overdamped
α>ω0
real, distinct roots
s1, s2
Underdamped
α<ω0
complex, conjugate
roots s1, s2*
Critically damped
α=ω0
real, equal roots
s1, s2
5
K. A. Saaifan, Jacobs University, Bremen
9.2 The Overdamped Parallel RLC Circuit
The condition of overdamped response ( 0 ) implies that
The roots of the characteristic equation s1 and s2 are distinct negative real
numbers
The response, v(t) , can be seen as a sum of two decreasing exponential
terms
st
s t
t∞
vt=A1 e A 2 e  0 as
1
2
Finding values for A1 and A2
For the shown circuit, we determine
1
1
0 =
=6
=
=3.5
2RC
 LC
s1 =−1
s2 =−6
The general form of the natural response
vt=A1 e−t A2 e−6 t
From the initial conditions v(0)=0 and iL(0)=-10 A
v0=A1 A2 =0
(1)
−A1−6 A2 =420
2
−I 0 −V 0 /R
d vC t
=
∣
dt t=0
C
=A1 s1 A2 s2
K. A. Saaifan, Jacobs University, Bremen
The final numerical solution is
vt=84 e−t −1e−6 t  V
Graphing the response
The maximum point can be determined as
d vt
=0
dt
−t
−6 t
=84−1 e −−61 e
=0
max
We determine the time t max =0.358 s
Then vtmax =48.9 V
max
K. A. Saaifan, Jacobs University, Bremen
Find an expression for vC(t) valid for t > 0 in the circuit
Compute the initial conditions (t < 0)
The capacitor acts as open circuits
200
vC 0−=150
=60 V
300200
The inductor acts as short circuits
−150
iL 0−=
=−300 mA
300200
After the switch is thrown (t > 0)
The capacitor is left in parallel with a 200 Ω
resistor and a 5 mH inductor
1
1
0 =
=100000
=
=125000
2RC
 LC
s1 =−50000
s2 =−200000
K. A. Saaifan, Jacobs University, Bremen
Solve the capacitor voltage
Since α > ω0, the circuit is overdamped and so
we expect a capacitor voltage of the form
vC t=A1 e−50000 t A2 e−200000 t
Finding values for A1 and A2
From the initial conditions vC(0)=60 V and iL(0)=-0.3 A
vC 0=A1 A2 =60
1
−50000A1−200000A2 =0
2
Solving, A1 = 80 V and A2 = −20 V, so that
vC t=80 e−50000 t −20 e−200000 t t0
−I 0 −V 0 / R
d vC t
=
dt ∣t=0
C
=A1 s1 A2 s2
9
K. A. Saaifan, Jacobs University, Bremen
(a) Sketch the voltage vR(t)= 2e−t−4e−3t V in the range 0<t<5 s
(b) Estimate the settling time
(c) Calculate the maximum positive value and the time at which it occurs
Graphing the response
The maximum point can be determined as
d vR t
=0
dt
−tmax
2−1e
−3 tmax
−4−3e
=0
ln 6
t max =
=0.895 s
2
vR tmax =544 mV
We compute the settling time as follows
v R tmax 
vR tsettling =
100
−tsettling
2e
−3 tsettling
−4e
=5.44 mV
−tsettling
2e
=5.44 mV
t settling=5.9 s
10
K. A. Saaifan, Jacobs University, Bremen
9.3 Critical Damping
The condition of a critical damping ( =0 ) implies that
The roots of the characteristic equation s1 and s2 are equal and
negative real numbers
s1=s2 =−
For repeated roots, the response, v(t) , can be seen as
vt=A1 te−t A2 e− t
Finding values for A1 and A2
For the shown circuit, we determine
0 ==  6 s−1
s1 =s2 =− 6 s−1
The general form of the natural response
vt=A1 te− 6 t A2 e− 6 t
From the initial conditions v(0)=0 and iL(0)=-10 A
v0=A2 =0
A1 − A2 =
10
=420
C
−I 0 −V 0 / R
d vC t
=
dt ∣t=0
C
=A1 − A2
K. A. Saaifan, Jacobs University, Bremen
The solution is
vt=420 te− 6 t V
Graphing the response
The maximum point can be determined as
d vt
=0
dt
=420e− t max420 tmax −e−t =0
We determine the time tmax
420e− t max 1− tmax =0
1
t max = =0.408 s

Then vt max =63.1 V
The settling time
vtmax 
− 6t
=420tsettling e 
100
settling
11
K. A. Saaifan, Jacobs University, Bremen
Find R1 such that the circuit is critically damped for t>0 and R2 so that v(0)=2 V
For t < 0
The capacitor acts as open circuits
The inductor acts as short circuits
v0−=5
R1 R2
=2 V
R1 R 2
After the switch is thrown (t > 0)
The current source has turned itself off and R2 is shorted
The capacitor is left in parallel with R1 and a 4 H inductor
1
1
1
=
=
0 =
=15810
2R 1 C 2 R1 ×10−9
 LC
Since the critically damping implies that 0 = , we have
R1 =31625 
31625 R 2
=0.4
31625R 2
R 2 =0.4 
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K. A. Saaifan, Jacobs University, Bremen
9.4 The Underdamped Parallel RLC Circuit
The condition of a critical damping ( 0 ) implies that
The roots of the characteristic equation s1 and s2 are complex conjugate
numbers
s =−j 
s =−  2−2
1

s =−−   −
1
d
0
2
s2 =−−j d
2
2
0
where d=  20 − 2 is the natural resonant frequency
For complex conjugate roots, the response, v(t) , can be seen as
vt=e− t A1 ej  t A2 e−j  t 
d
d
=e−t [A1 cos d tj sin d tA2 cos d t−j sin d t]
=e−t [A1 A2 cos d tj A1−A2 sin d t]
=e−t [B 1 cos d tB 2 sin d t]
The derivative of v(t) is
dvt
=−e−t [B1 cos d tB 2 sin d t]
dt
−t
e [B 1 −d sin d tB 2 d cos d t]
=e−t [− B1 d B2  cosd t− B2 d B1  sin d t]
K. A. Saaifan, Jacobs University, Bremen
14
Finding values for B1 and B2
For the shown circuit, we determine
1
1
=
=2 s−1
0 =
=  6 s−1
2RC
 LC
d=  20 − 2=  2 s−1
The natural response is
vt=e−2 t [B 1 cos  2tB2 sin  2 t]
From the initial conditions v(0)=0 and iL(0)=-10 A
v0=B1 =0
 2 B 2=420
The final numerical solution is
vt=210  2 e−2 t sin  2 t
−I 0 −V 0 / R
d vC t
=
dt ∣t=0
C
=− B1d B 2
K. A. Saaifan, Jacobs University, Bremen
Graphing the response
vt=210  2 e−2 t sin  2 t
The voltage oscillates (~ωd) and approaches to the final value (~α)
The voltage response has two extreme points (minimum and maximum
points)
15
K. A. Saaifan, Jacobs University, Bremen
Find an expression for vC(t) valid for t > 0 in the circuit
Compute the initial conditions (t < 0)
The capacitor acts as open circuits
48×100
vC 0−=3
=97.3 V
10048
The inductor acts as short circuits
100
iL 0−=3
=2.027 A
48100
After the switch is thrown (t > 0)
The current source is off
The capacitor is left in parallel with a 48 Ω resistor and a 10 H inductor
1
0 =
=4.99 s−1 = 1 =1.2 s−1
 LC
2RC
Since 0 , the circuit is underdamped
vC t=e− t [B1 cos d tB 2 sin d t]
2
2
−1
where d=  0 − =4.75 s
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17
K. A. Saaifan, Jacobs University, Bremen
Finding values for B1 and B2
The natural response is
vC t=e−1.2 t [B1 cos 4.75 tB 2 sin 4.75 t]
From the initial conditions vC(0)=97.3 and iL(0)=2.027 A
v0=B1 =97.3
4.75B 2=240−2.027−97.3/100
The final numerical solution is
−I 0 −V 0 / R
d vC t
=
dt ∣t=0
C
=− B1d B 2
vC t=e−1.2 t [97.3 cos 4.75 t−151.57 sin 4.75 t] V
18
K. A. Saaifan, Jacobs University, Bremen
9.5 The Source-Free Series Circuit
Obtaining the differential equation of a series RLC circuit
Apply KVL (V0, I0, i(t) must satisfy the passive sign convention)
R ivL tvC t=0
di 1 t
R iL  ∫t i d t vC t0 =0
dt C
0
Differentiate both sides with respect to time
d2 i
di 1
L 2 R
 i=0
d
t
C
dt
The two initial conditions
The inductor current cannot change abruptly
i0=I 0
1
The capacitor voltage cannot change abruptly
vC 0=V 0
−V 0−I 0 R
d iL t
=
dt ∣t=0
L
2
vL 0=−vC 0−vR 0
=−V 0 −I 0 R
vL 0=L
d iL t
dt ∣t=0
19
K. A. Saaifan, Jacobs University, Bremen
Solution of the differential equation
We assume the exponential form of the natural response is
vt=Aest
Substitute into the ordinary differential equation, we got the characteristic
equation of s determined by the circuit parameters
1
Ls2 R s =0
C
The characteristic equation has two roots
R
s1,2 =−
±
2L

R 2 1
−
LC
2L

=−±  2 −20
where
The resonant frequency
The damping coefficient
1
 LC
R
=
2L
0 =
20
K. A. Saaifan, Jacobs University, Bremen
Summary of Relevant Equations for Series Source-Free RLC Circuits
Condition
Overdamped
Underdamped
Critically damped
Criteria
α
ω0
α>ω0
R
2L
1
 LC
α<ω0
R
2L
1
 LC
α=ω0
R
2L
1
 LC
Response
s1 t
it=A1 e A2 e
where
s2 t
s1,2 =−±  2 −02
it=e−t [B 1 cos d tB 2 sin d t]
2
2
where d=  0 −
it=A1 te−t A2 e− t
K. A. Saaifan, Jacobs University, Bremen
9.6 The Complete Response of the RLC Circuit
The response of RLC circuits with dc sources and switches consists of the
natural response and the forced response:
Forced Response
Natural Response
v t= v


f t
Forced Response
it=
i t

f
v t

n
Natural Response

i t

n
The general solution is obtained by the same procedure that was followed for
RL and RC circuits
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22
K. A. Saaifan, Jacobs University, Bremen
The Solution Steps of RLC Circuits
Determine the initial conditions
Compute the circuit current, “iL(t), iR(t), and iC(t)”, and voltages,
“vL(t), vR(t), and vC(t)”, at t=0- and t=0+
“Note that the inductor current the capacitor voltage cannot change abruptly,
iL(0-)=iL(0)=iL(0+) and vC(0-)=vC(0)=vC(0+) ”
hf(t)
Upon we are confronted with a series or a parallel circuit
R
1
=
(series RLC)
=
(parallel RLC)
2L
2 RC
1
 0=
 LC
− t
hn t=A1 te
−t
=0
0
s1 t
hn t=A1 e A2 e
A2 e
d=  20 −2
−t
hn t=e
0
s1,2 =−±  2 −02
The forced response
s2 t
The complete response
hf(t)+hn(t)
[B 1 cos d tB 2 sin d t]
Find unknown constants
given the initial conditions
23
K. A. Saaifan, Jacobs University, Bremen
Find an expression for vC(t) and iL(t)
valid for t > 0 in the circuit
-
+
1. Determine the forced response ( t > 0 )
iLf t=−9 A
t0
vCf t=150 V
t0
2. Determine the natural response
-
1. Write the differential equation
R ivL tv C t=0
+
The differential equation in terms of i reduces to
2
d i
di 1
L 2 R
 i=0
d
t
C
dt
Note:
Independent current sources → open circuits
Independent voltage sources → short circuits
The differential equation in terms of vC(t) is given as
2
L
d vC
2
dt
R
d vC 1
 v =0
dt C C
2. Compute αand ω0
1
−1
0=
=3 s
 LC
s1 =−1
R
−1
=
=5 s
2L
s2 =−9
3. Since α>ω0, the response is overdamped, we have
and
−t
−9 t
−t
−9 t
iLn t=A1 e A2 e
vC n t=B 1 e B 2 e
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K. A. Saaifan, Jacobs University, Bremen
3. The complete response
vC t=vCf tB1 e−tB2 e−9 t
iL t=iLf tA1 e−t A2 e−9 t
=−9A1 e−t A 2 e−9 t
=150B1 e−tB2 e−9 t
4. Solve for the values of the unknown constants
vC t ∣t=0 =vC 0 
150B1 B 2 =150
1
-
4
+

dv C t
iC 0 
=
dt ∣t=0
C
−B 1−9 B2 =108
For t=0- (the left-hand current source is off)
2
−t
vC t=15013.5 B1 e −13.5B 2 e
−9 t
iL t ∣t=0 =iL 0 
−9A1 A2=−5
diL t
vL 0
=
dt ∣t=0
L
−A1 −9 A2 =−40
iL t=−9−0.5 e−t 4.5 e−9 t
1
−
vL 0 =0 V
−
vR 0 =−150 V
−
vC 0 =150 V
iL 0 =−5 A
iR 0 =−5 A
iC 0 =0 A
−
−
For t=0+ (the left-hand current source is on)

vL 0 =−120 V

vR 0 =−30 V

vC 0 =150 V
iL 0 =−5 A
iR 0 =−1 A
iC 0 =4 A
2
−



K. A. Saaifan, Jacobs University, Bremen
9.7 THE LOSSLESS LC CIRCUIT
The resistor in the RLC circuit serves to dissipate initial stored energy
When this resistor becomes 0 in the series RLC or infinite in the parallel RLC,
the circuit will oscillate
Example: Assume the shown circuit with the
following initial conditions
1
vC 0=0 V
iL  0=− A
6
1. We find
1
R
−1
−1
0 =
=3 s
=
=0 s
 LC
2L
−1
2. So d=3 s
, the voltage is simply
− t
vt=e
[B 1 cos 3 tB2 sin 3 t]
3. We use the initial condition
v0=B1
B 1=0
dvC t
−iL 0 
B 2=2
=
=3 B 2
∣
dt t=0
C
4. Thus, we have obtained a sinusoidal response
vt=2 sin 3 t V
25
K. A. Saaifan, Jacobs University, Bremen
Homework Assignment 8
P9.1, P9.6, P9.12, P9.13, P9.16, P9.20, P9.26, P9.27, P9.35, P9.37, P9.46
P9.50, P9.51, and 9.64
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